P(X < 14.0) = P((x - 15.7) / 0.7) < (14.0 - 15.7) / 0.7) = P(Z < -2.43) = 0.0076
P(X < 18.0) = P((x - 15.7) / 0.7) < (18.0 - 15.7) / 0.7) = P(Z < 3.29) = 0.9995
or use calc by typing in normalcdf(-99,-2.43) and normalcdf(-99,3.29)you get the same probabilities of .0076 and .9995
P(14.0 < x < 18.0) = 0.9995 - 0.0076 = .9919 have their shirts in stock, 1-.9919 = .0081 do not have shirts in stock
Mens shirt sizes are determined by neck sizes. suppose the mens nexck sizes are approximatley normally distributed with mean 15.7 inches and standard deviation 0.7 inch.retailer sells men shirts in S,M,L,XL where the shirt sizes are defined in the table below.
S: 14 ≤ neck size <15
M: 15 ≤ neck size <16
L: 16 ≤ neck size <17
XL:17 ≤ neck size <18
because the retailer only stocks the sizes listed above what proportion of customers will find the retailer does not carry any shirts in their size?
1 answer
1 answer